How big the uBeam mouse will be.... Simple enough...
The transmitter sends out a collimated beam of 1KW/M^2, e.g. 0.1W per cm^2 of optical power.
There's some range, which means about 50% loss (Air is lossy), and conversion of ultrasound to electricity can be done at some 10% efficiency (commercial 1-2%)
Recievers have to work at some angles, I believe 45 degrees is minimal, so you need to reduce this by 30% more.
So that's about 3mW per cm^2 receiver, or 20mW/Inch^2
A mouse consumes about 1mW of power to run it and process the data, so the receiver can be small enough.
Now do the same for a phone.... A phone uses 5W to charge.... so you will need some 250 Inch^2 to charge it, you will need a 20" screen phablets (or phaTVs?) will easily accomodate it.
Not sure how you get that number for the phone. If at 150 dB that needs 0.005m^2 to get 5W, let's be super generous and say 10W with 50% efficiency of receive conversion, so 0.01m. An iPhone X is about 7.5 by 15cm so 0.01125m^2.
Wow, isn't that lucky. 150 dB works out perfectly, under perfect conditions and a receive efficiency similar to what you see in one of those pitch graphs, to charge an iPhone X rapidly. So if conditions aren't perfect, or 150dB is deemed unsafe, or receive efficiency can't approach 50%, then it's downhill from there.
Oh, and that's from a 0.33m^2 transmitter, say at 50% efficiency, so 666W used, so <1% efficiency.
Unlike you, I'm not an ultrasound expert, so....
But I'm a practical person.... and believe products should be designed to be working, not to be lab experiments published in a research paper.
For a phone to charge, you need 5W of power, in normal conditions, that are actually competitive to available solutions, e.g. better compared to a Qi pad or usb cable.
The way I interpret the above is working when tilted to some degree (45 degrees sounds reasonable, although I would prefer 60-70)
At some distance (say 5-10 feet)
Even at 100% humidity, and cold/hot weather.
A phone that's as clean as a normal phone is, possibly with some of the area obscured.
It has to be safe, legal, and not annoying (to me, my kids or my pets, also, for it to be in my livingroom, it should not make a fan noise)
the number 1KW/M^2 is taken from uBeam's ppt, it's comparable to 150dB
I know it's possible to focus that power to a smaller spot with a phased array, but considering uBeam must be transmitting at a safe level (I'll take their word 150dB is safe and legal, althogh I know it to be not true), the beam cannot exceed this level at any point, in any weather conditions. So, if attenuation for dry air for 10-20dB higher compared to moist air, then the power density at the receiver can't exceed ~140dB acoustic.
Taking into account your 50% conversion efficiency (I believe you, although I never seen anything close) that's equivalent to 5mW/cm^2
The effective area of iPhone X tilted at 45 degrees is 70 cm^2 - that's not enough (if it were enough a solar cell would be enough to keep the phone charged, no transmitter needed).
at 5mW/cm^2 you need 1000 cm^2, taking 45 degrees phone tilt into account you need 1400cm^2
Considering phones are ~1-2 length/width ratio, that's ~35cm X 70 cm phone.
The diagonal of the screen would be almost 31 Inch.
If you're willing to charge a bit slower on a humid day.... I think 20 Inch would be enough.